{"id":35256,"date":"2024-12-20T13:00:15","date_gmt":"2024-12-20T13:00:15","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35256"},"modified":"2025-12-11T17:06:13","modified_gmt":"2025-12-11T17:06:13","slug":"weierstrass-extreme-value-theorem","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/","title":{"rendered":"Weierstrass Extreme Value Theorem"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Weierstrass Extreme Value Theorem<\/h1>\n<p style=\"text-align:center;\"><em>Why is it that in so many optimization problems it is almost taken for granted that \u201cthe maximum exists\u201d or that \u201cthere is always a minimum\u201d on a given interval, when in reality nothing forces that to be the case? The <strong>Weierstrass Theorem<\/strong> is the missing piece of that puzzle: it guarantees that a continuous function defined on a closed and bounded interval not only is bounded, but also actually attains its extreme values. In this entry we review its statement, build a detailed and rigorous proof based on pointwise continuity, compactness, and the supremum axiom, and comment on its modern interpretation in terms of continuous functions on compact sets. The aim is that, by the end, you do not merely recall the theorem as a sentence, but understand why it is true and why it appears repeatedly in analysis, optimization, and applied models.<\/em><\/p>\n<p style=\"text-align:center;\"><b>Learning Objectives<\/b><\/p>\n<ol>\n<li>\n    <strong>Understand the statement of the Weierstrass Theorem.<\/strong><br \/>\n    Identify precisely the hypotheses of the theorem (continuous function on a closed and bounded interval <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span>) and its main conclusions: boundedness and the existence of maximum and minimum values.\n  <\/li>\n<li>\n    <strong>Interpret the Weierstrass Theorem in terms of compactness.<\/strong><br \/>\n    Formulate the result in modern language: continuous functions map compact sets into sets where extreme values are attained, connecting the case of <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span> with the broader framework of real analysis.\n  <\/li>\n<li>\n    <strong>Relate the Weierstrass Theorem to optimization problems.<\/strong><br \/>\n    Recognize the role of the theorem as the theoretical foundation for the existence of maxima and minima in many single-variable optimization problems, in both theoretical and applied contexts.\n  <\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>TABLE OF CONTENTS<\/u>:<\/b><br \/>\n<a href=\"#1\"><b>Introduction<\/b><\/a><br \/>\n<a href=\"#2\"><b>Statement of the Weierstrass Theorem<\/b><\/a><br \/>\n<a href=\"#3\">Proof<\/a><br \/>\n<a href=\"#4\">Step 1: Pointwise continuity on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#5\">Step 2: Open cover associated with continuity<\/a><br \/>\n<a href=\"#6\">Step 3: Compactness of <span dir=\"ltr\">[a,b]<\/span> and finite subcover<\/a><br \/>\n<a href=\"#7\">Step 4: Construction of a <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> independent of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (uniform continuity)<\/a><br \/>\n<a href=\"#8\">Step 5: From uniform continuity to boundedness of <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#9\">Step 6: Existence of maximum and minimum values<\/a><br \/>\n<a href=\"#10\"><b>Interpretation in terms of compactness and conclusion<\/b><\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/N5mSrhJgCds\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/br><\/p>\n<h2>Introduction<\/h2>\n<p>\nThe <strong>Weierstrass Extreme Value Theorem<\/strong> is one of those results that, although it usually appears in the early units of Real Analysis, in fact silently supports a large portion of applied mathematics. Whenever in physics, economics, or statistics we speak of \u201cmaximizing\u201d or \u201cminimizing\u201d a quantity subject to certain constraints, we are essentially relying on an idea very close to what this theorem guarantees: that a continuous function defined on a closed and bounded interval <strong>is not only bounded, but actually attains its extreme values<\/strong>.\n<\/p>\n<p>\nIntuitively, it may seem \u201cobvious\u201d that if we draw a continuous curve over a segment <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, then there must exist a highest and a lowest point. However, it is enough to make small changes in the hypotheses for this intuition to fail dramatically: if we open the interval, if the function ceases to be continuous, or if the domain is not bounded, maxima and minima may simply disappear. The Weierstrass Theorem brings order to this intuition and tells us precisely <em>when<\/em> we can rely on it and <em>why<\/em>.\n<\/p>\n<p>\nFrom a theoretical perspective, this theorem is the first serious encounter with the idea of <strong>compactness<\/strong>: in modern language, what it states is that a continuous function maps compact sets into compact sets. From a practical perspective, this translates into the existence of solutions for many one-dimensional optimization problems, and it will be a key component for later results such as the <b>Mean Value Theorem<\/b> and, ultimately, for calmly understanding the Fundamental Theorem of Calculus.\n<\/p>\n<p>\nIn this section we will state the Weierstrass Theorem and develop its proof in detail, relying on the notion of continuity on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> and on the supremum axiom. The idea is that this text serves as a solid reference: both for studying the result itself and for returning to it whenever you need to use it in proving other theorems or in rigorously justifying the existence of maxima and minima in concrete problems.\n<\/p>\n<p><a name=\"2\"><\/a><\/br><\/p>\n<h2>Statement of the Weierstrass Theorem<\/h2>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: justify; background-color: #e0e0ff;\">\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=N5mSrhJgCds&amp;t=439s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Every function <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> defined<\/span><\/strong><\/a> and continuous on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b],<\/span><\/span> is bounded and has minimum and maximum values, <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">M<\/span>, such that if <span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span>, then <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)\\in[m,M]<\/span><\/span>.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"3\"><\/a><\/br><\/p>\n<h3>Proof<\/h3>\n<p>\nLet us prove that if <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> is continuous on the closed and bounded interval <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, then <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is bounded and attains a maximum and a minimum value on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. We will divide the proof into two major parts:\n<\/p>\n<ul>\n<li>First, we will show that the continuity of <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> implies that <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is <em>uniformly continuous<\/em>, and from this we will deduce that it is <strong>bounded<\/strong>.<\/li>\n<li>Then, using the supremum axiom, we will prove that <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> attains its maximum and minimum values on the interval.<\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/br><\/p>\n<h4><b>Step 1:<\/b> Pointwise continuity on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nBy hypothesis, <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is continuous at every point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>. By the definition of continuity in terms of <span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span>, this means that:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall \\epsilon\\gt 0)(\\exists \\delta(x_0)\\gt 0)\n\n\\big(|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\nAt this point, the number <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong> may depend on the point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>. Our immediate goal will be to construct, from these <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong>, a single number <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> that does not depend on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> and that works simultaneously for all points in the interval.\n<\/p>\n<p><a name=\"5\"><\/a><\/br><\/p>\n<h4><b>Step 2:<\/b> Open cover associated with continuity<\/h4>\n<p>\nFix an arbitrary <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\\gt 0<\/span><\/span><\/strong>. For each <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, the continuity of <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> allows us to choose a number <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)\\gt 0<\/span><\/span><\/strong> such that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\nFrom these values we define, for each <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, an open interval\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_0}=\\left(x_0-\\frac{\\delta(x_0)}{2},\\,x_0+\\frac{\\delta(x_0)}{2}\\right).\n\n<\/span>\n<\/p>\n<p>\nEach <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong> is an open set in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span> and, moreover, the family\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{I_{x_0}\\}_{x_0\\in[a,b]}\n\n<\/span>\n<\/p>\n<p>\nforms an <strong>open cover<\/strong> of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Indeed, given any point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in[a,b]<\/span><\/span><\/strong>, it suffices to take <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0=y<\/span><\/span><\/strong>; by construction, <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in I_y<\/span><\/span><\/strong>. Thus, every point of the interval belongs to at least one of the open sets <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong>.\n<\/p>\n<p>\nThis family of open sets is, in general, <strong>infinite<\/strong> (there is one for each <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>). This is where the compactness of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> comes into play.\n<\/p>\n<p><a name=\"6\"><\/a><\/br><\/p>\n<h4><b>Step 3:<\/b> Compactness of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span> and finite subcover<\/h4>\n<p>\nWe know from the Heine\u2013Borel Theorem that a subset of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span> is compact if and only if it is closed and bounded. The interval <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> is closed and bounded, therefore it is compact. By the definition of compactness, this means that:\n<\/p>\n<p>\nFrom <strong>every<\/strong> open cover of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> (even if it has infinitely many sets) one can extract a <strong>finite subcover<\/strong>.\n<\/p>\n<p>\nApplying this property to the open cover <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{I_{x_0}\\}_{x_0\\in[a,b]}<\/span><\/span><\/strong>, it follows that there exist points <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1,\\dots,x_N\\in[a,b]<\/span><\/span><\/strong> such that the corresponding intervals\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_1},\\, I_{x_2},\\,\\dots,\\,I_{x_N}\n\n<\/span>\n<\/p>\n<p>\nthey still cover the entire interval:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n[a,b]\\subset I_{x_1}\\cup I_{x_2}\\cup\\cdots\\cup I_{x_N}.\n\n<\/span>\n<\/p>\n<p>\nThus, we have gone from an infinite family of open intervals to a subcover consisting of only a <strong>finite number<\/strong> of intervals, without losing the property of covering <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"7\"><\/a><\/br><\/p>\n<h4><b>Step 4:<\/b> Construction of a <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> that does not depend on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (uniform continuity)<\/h4>\n<p>\nFrom the finite subcover we define the number\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\delta=\\min\\left\\{\\frac{\\delta(x_1)}{2},\\frac{\\delta(x_2)}{2},\\dots,\\frac{\\delta(x_N)}{2}\\right\\}.\n\n<\/span>\n<\/p>\n<p>\nSince this is the minimum of a finite collection of positive numbers, we have <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>. We will show that this <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> works for <strong>every<\/strong> point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, meaning it does not depend on the choice of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>.\n<\/p>\n<p>\nNow take:\n<\/p>\n<ul>\n<li>an arbitrary point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, and<\/li>\n<li>a point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> such that <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong>.<\/li>\n<\/ul>\n<p>\nSince the intervals <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_1},\\dots,I_{x_N}<\/span><\/span><\/strong> cover <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, the point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> belongs to at least one of them, say to <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong> for some <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">j\\in\\{1,\\dots,N\\}<\/span><\/span><\/strong>. By the definition of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong>, this means that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x_0-x_j|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nMoreover, by the definition of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/span><\/strong> we have <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\le\\frac{\\delta(x_j)}{2}<\/span><\/span><\/strong>, so from <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong> it follows that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nApplying the triangle inequality,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_j|\\le |x-x_0|+|x_0-x_j|\n\n\\lt \\frac{\\delta(x_j)}{2}+\\frac{\\delta(x_j)}{2}\n\n=\\delta(x_j).\n\n<\/span>\n<\/p>\n<p>\nBy the choice of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_j)<\/span><\/span><\/strong> (continuity of <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> at <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_j<\/span><\/span><\/strong> for the value <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\/2<\/span><\/span><\/strong>), the inequalities <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x_0-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> and <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> imply\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x_0)-f(x_j)|\\lt\\frac{\\epsilon}{2}\n\n\\quad\\text{and}\\quad\n\n|f(x)-f(x_j)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\nUsing again the triangle inequality we obtain\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_0)|\n\n\\le |f(x)-f(x_j)| + |f(x_j)-f(x_0)|\n\n\\lt \\frac{\\epsilon}{2}+\\frac{\\epsilon}{2}\n\n=\\epsilon.\n\n<\/span>\n<\/p>\n<p>\nSince <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> and <strong><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/strong> were arbitrary, we have shown that for the <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span><\/strong> chosen at the beginning there exists a <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>, independent of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>, such that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall x\\in[a,b])\n\n\\big(|x-x_0|\\lt\\delta\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\nIf we rename <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> as <strong><span class=\"katex-eq\" data-katex-display=\"false\">y<\/span><\/strong>, this can be written as:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall \\epsilon\\gt 0)(\\exists \\delta\\gt 0)(\\forall x,y\\in[a,b])\n\n\\big(|x-y|\\lt\\delta\\Rightarrow |f(x)-f(y)|\\lt\\epsilon\\big),\n\n<\/span>\n<\/p>\n<p>\nwhich is precisely the definition of <strong>uniform continuity<\/strong> of <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. In what follows, we will only need to apply this result to the case <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"8\"><\/a><\/br><\/p>\n<h4><b>Step 5:<\/b> From uniform continuity to boundedness of <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> on <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nLet us now apply uniform continuity with <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>. There exists a number <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1\\gt 0<\/span><\/span><\/strong> such that for all <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x,y\\in[a,b]<\/span><\/span><\/strong> we have\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-y|\\lt\\delta_1\\Rightarrow |f(x)-f(y)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nWe now divide the interval <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> into a finite number of subintervals whose length is smaller than <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1<\/span><\/span><\/strong>. That is, we choose an integer <strong><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/strong> and points\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\na = x_0 \\lt x_1 \\lt \\cdots \\lt x_n = b\n\n<\/span>\n<\/p>\n<p>\nsuch that for each <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,1,\\dots,n-1<\/span><\/span><\/strong> we have\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nx_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nWe now consider the finite set of values\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{f(x_0),f(x_1),\\dots,f(x_{n-1})\\}.\n\n<\/span>\n<\/p>\n<p>\nBeing a finite set of real numbers, we can define without issue\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nC = \\max\\{|f(x_k)| \\;|\\; k=0,1,\\dots,n-1\\}.\n\n<\/span>\n<\/p>\n<p>\nWe will now show that <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C+1<\/span><\/span><\/strong> is an upper bound in absolute value for <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> on the entire interval <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Let <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> be an arbitrary point. Then there exists an index <strong><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/strong> such that <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[x_k,x_{k+1}]<\/span><\/span><\/strong>. In particular,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_k|\\le x_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nBy uniform continuity with <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>, from <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_k|\\lt\\delta_1<\/span><\/span><\/strong> it follows that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_k)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nUsing the triangle inequality:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le |f(x)-f(x_k)| + |f(x_k)| \\lt 1 + |f(x_k)| \\le 1 + C.\n\n<\/span>\n<\/p>\n<p>\nSince <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> was arbitrary, we conclude that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le C+1 \\quad \\text{for all } x\\in[a,b],\n\n<\/span>\n<\/p>\n<p>\nthat is, the function <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is <strong>bounded<\/strong> on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"9\"><\/a><\/br><\/p>\n<h4><b>Step 6:<\/b> Existence of maximum and minimum values<\/h4>\n<p>\nWe define the set of values taken by the function on the interval:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nH=\\{f(x)\\;|\\;x\\in[a,b]\\}\\subset\\mathbb{R}.\n\n<\/span>\n<\/p>\n<p>\nWe already know that <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> is nonempty (because <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> is nonempty) and bounded, so by the supremum axiom there exist real numbers\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM=\\sup H,\\qquad m=\\inf H.\n\n<\/span>\n<\/p>\n<p>\nLet us prove that <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> is attained as a value of the function, that is, that there exists <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> with <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x_1)=M<\/span><\/span><\/strong>. We proceed by contradiction.\n<\/p>\n<p>\nSuppose that <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> never attains the value <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong>, that is:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(f(x)\\lt M\\big).\n\n<\/span>\n<\/p>\n<p>\nUnder this assumption, the function\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\ng(x)=\\frac{1}{M-f(x)}\n\n<\/span>\n<\/p>\n<p>\nis well defined and positive for all <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong>, since by hypothesis <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-f(x)\\gt 0<\/span><\/span><\/strong>. Moreover, since <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is continuous and <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> is constant, <strong><span class=\"katex-eq\" data-katex-display=\"false\">g<\/span><\/strong> is also continuous. By the first part of the proof, every continuous function on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> is bounded, so there exists a number <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">N\\gt 0<\/span><\/span><\/strong> such that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(g(x)\\le N\\big).\n\n<\/span>\n<\/p>\n<p>\nIn particular, for all <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> we have\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\frac{1}{M-f(x)} = g(x)\\le N,\n\n<\/span>\n<\/p>\n<p>\nwhich is equivalent to\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM-f(x)\\ge \\frac{1}{N}\n\n\\quad\\Rightarrow\\quad\n\nf(x)\\le M-\\frac{1}{N}.\n\n<\/span>\n<\/p>\n<p>\nThis means that all values of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> are less than or equal to <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-\\frac{1}{N}<\/span><\/span><\/strong>. In particular, the supremum of <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> satisfies\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\sup H\\le M-\\frac{1}{N}\\lt M,\n\n<\/span>\n<\/p>\n<p>\nwhich contradicts the definition of <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> as the supremum of <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong>. Therefore, our initial assumption was false, and there must exist a point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> such that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_1)=M.\n\n<\/span>\n<\/p>\n<p>\nA completely analogous argument, applied to the infimum <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=\\inf H<\/span><\/span><\/strong> (for example, by considering the function <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">h(x)=-f(x)<\/span><\/span><\/strong>), shows that there exists a point <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2\\in[a,b]<\/span><\/span><\/strong> such that\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_2)=m.\n\n<\/span>\n<\/p>\n<p><a name=\"10\"><\/a><\/br><\/p>\n<h2>Interpretation in terms of compactness and conclusion<\/h2>\n<p>\nWe have proved that every continuous function <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> is bounded and attains its maximum and minimum values on <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. In the modern language of analysis, this is interpreted by saying that in <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>, closed and bounded intervals such as <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> are compact sets, and continuous functions map compact sets into compact sets.\n<\/p>\n<p>\nIn particular, if <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong> is compact and <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> is continuous on <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong>, then the image <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> is a compact subset of <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>. This guarantees that <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> is bounded and that it actually attains a maximum and a minimum value, which is precisely the content of the Weierstrass Theorem.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Weierstrass Extreme Value Theorem Why is it that in so many optimization problems it is almost taken for granted that \u201cthe maximum exists\u201d or that \u201cthere is always a minimum\u201d on a given interval, when in reality nothing forces that to be the case? The Weierstrass Theorem is the missing piece of that puzzle: it [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35255,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":6,"footnotes":""},"categories":[854,567],"tags":[],"class_list":["post-35256","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-differential-calculus","category-mathematics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Weierstrass Extreme Value Theorem - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Understand the Weierstrass Theorem from scratch: continuity, compactness, and a step-by-step proof applied to maximum and minimum problems.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Weierstrass Extreme Value Theorem\" \/>\n<meta property=\"og:description\" content=\"Understand the Weierstrass Theorem from scratch: continuity, compactness, and a step-by-step proof applied to maximum and minimum problems.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2024-12-20T13:00:15+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-12-11T17:06:13+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1-1024x683.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Weierstrass Extreme Value Theorem\" \/>\n<meta name=\"twitter:description\" content=\"Understand the Weierstrass Theorem from scratch: continuity, compactness, and a step-by-step proof applied to maximum and minimum problems.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"10 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Weierstrass Extreme Value Theorem\",\"datePublished\":\"2024-12-20T13:00:15+00:00\",\"dateModified\":\"2025-12-11T17:06:13+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\"},\"wordCount\":2693,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\",\"articleSection\":[\"Differential Calculus\",\"Mathematics\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\",\"url\":\"https:\/\/toposuranos.com\/material\/en\/weierstrass-extreme-value-theorem\/\",\"name\":\"Weierstrass Extreme Value Theorem - 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